A cognitive retraining clinic assists outpatient victims of

A cognitive retraining clinic assists outpatient victims of head injury, anoxia, or other conditions that result in cognitive impairment. Each incoming patient is evaluated to establish an appropriate treatment program and estimated length of stay. To see if the evaluation teams are consistent, 12 randomly chosen patients are separately evaluated by two expert teams (A and B) as shown. The data file is named LengthStay. Estimated Length of Stay in Weeks At the .10 level of significance, is there a difference between the evaluator teams\' estimated length of stay? State your hypotheses and show all 7 steps clearly. Give and interpret the p-value. What assumption is required for this test? Use an appropriate graph to check the assumption and state your findings. Verify with Minitab, but attaching or including the output.

Solution

Step 1: Definition of Hypothesis

Significance Level = 0.1 = Alpha

Confidence Interval = 1 - Alpha = 0.9

Null Hypothesis: Difference between mean of Team A and B is zero.

Alternative Hypothesis: Difference between mean of Team A and B is not zero.

Step 2: Computation of Difference

Mean of Difference (dBar) is -2.5

Sample standard deviation of difference(s) is 4.9082

Sample size(n) is 12.

Step 3: Calculation of Statistic Parameters

In this problem, degree of freedom (df) = 11

For df=11, alpha=0.1 and two tailed test, corresponding t value is +/- 1.796 [Looked up from t-table]

Formula for t-statistic is dBar - D / ( s / sqrt(n) )

D = Population Difference = 0 (Assumed in this problem)

t = -2.5 / (4.9082 / sqrt(12) ) = -1.76444

Step 4: Acceptance or rejection using t-statistic

Since t-statistic lies in the range (-1.796,1.796), Null hypothesis cannot be rejected.

b ) Calculation of p-value

Since we are conducting two tailed test, p - value should be calculated as 2 * P(t < -1.7644)

-1.7644 is the t-statistic value calculated.

P-Value = 2 * P(t < -1.7644) = 2*0.0526 = 0.1053

P-Value is slightly higher than significance level alpha. Hence, Null hypothesis can\'t be rejected.

c) Assumed that values are normally distributed

d) Minitab is a paid software, hence unable to use it. Can support with R output if necessary.

Team/Patient	1	2	3	4	5	6	7	8	9	10	11	12
A	24	24	52	30	40	30	18	30	18	40	24	12
B	24	20	52	36	36	36	24	36	16	52	24	16
Difference	0	4	0	-6	4	-6	-6	-6	2	-12	0	-4